Chapter 25 Electromagnetic WavesChapter 25 Electromagnetic Waves

Outline

25-1 The production of Electromagnetic Waves

25-3 The Electromagnetic Spectrum

25-4 Energy and momentum in Electromagnetic Wavesgy g

25-5 Polarization

25-1 The production of Electromagnetic Waves

Electromagnetic waves: Visible light (440 ~ 700nm) is a kind of electromagnetic waves

Electromagnetic waves are transverse waves: the vibration of each point is vertical to the propagation direction.

An Electromagnetic wave consists of two vertical components: (electric filed) and (Magnetic field)E Bcomponents: (electric filed) and (Magnetic field).E

BThe electric field vector , magnetic field , and the propagation E

B

, g , p p gare all vertical each other !

The Electromagnetic waves travel through vacuum (air) at the same speed:same speed:

C = 3.00 × 108 m/s, the speed of light !

Figure 25-3An Electromagnetic WaveAn Electromagnetic Wave

at an instant moment t.

Direction of Propagation for Electromagnetic Waves

Point the fingers of your right hand in the direction of the , curl Eg y g ,your fingers toward , and your thumb will point in the direction of the propagation.

B

25-3 Electromagnetic Spectrum

When white light passes through a prism, it spread out into a rainbow of different colors. This indicates that the electromagnetic waves have different wavelengths, since different colors have different wavelength.

The wavelength and frequency is associated each other as,The wavelength and frequency is associated each other as,

)425( −= λfc

Where c is the speed of the electromagnetic wavelength in vacuum and is a constant. C is in m/s.

λ Is in m

f is in Hz.f

Wavelength λ, frequency f, and propagation direction.

Example 25-3a p e 5 3

Find the frequency of red light with a wavelength of 700.0 nm.

Solution

Since c=fλ, we have,

8 9

8

3.00 10 / (700.0 10 )3 00 10 /

m s f mm s

−× = ×

× 149

3.00 10 / 4.29 10700.0 10

m sf Hzm−

×= = ×

×

Example 25-3The spectrum of white light.

Figure 25-8The ElectromagneticThe Electromagnetic

Spectrum

25-4 Energy and momentum in Electromagnetic Waves

Since energy density (energy per volume) of electric field (Section 20-6) is,

21 Eu ε=

And since energy density of magnetic field (Section 23-9) is,

02EuE ε=

2

021 BuB μ

=0

As the electromagnetic wave is composed of electric and magnetic fields, the total energy density of the electromagnetic wave is

)525(21

21 22

0 −+=+= BEuuu EB ε )(22 0

0EB μ

It can be shown that electric and magnetic energy densities are l S hequal. So, one has

)625(121

21 22

022

0 −==+= BEBEu εε )(22 0

00

0 μμ

Since E and B are sinusoidal functions of the time, we must use RMS value to calculate the “average” valueRMS value to calculate the average value,

)725(121

21 22

022

0 −==+= rmsrmsrmsrmsav BEBEu εε )(22 0

00

0 rmsrmsrmsrmsav μμ

Where,

2max=

EErms

)825(2

max −=BBrms

The relationship of E and B

Since,2 2

00

1 , (25 6)E B fromεμ

= −

We have BE00

1με

=

12 2 20

70

8.85 10 /( ) (19 12)

4 10 / (22 8)

Pemittivity of free space C N m

Permeability of free space T m A

ε

μ π

−

−

= × ⋅ −

= × ⋅ −

80 01/ 3.00 10 /c m sε μ= = ×

so we eventually have

(25 9)E cB= −( )rms rmsE cB=

Exercise 25-3

At a given instant of time, the electric field in a beam of sunlight has a magnitude of 510 N/C. What is the magnitude of the magnetic field at this instant?

E BcSolution:

8510 / (3.00 10 / )510 /

E BcN C B m sE N C

=

= ×

68

510 / 1.7 103.00 10 /

E N CB Tc m s

−= = = ××

Intensity IThe amount of the energy a wave delivers to a unit area in a unit time is

In figure 25-9, the intensity I is

gycalled Intensity I. Also I = P/A

uctAtAcu

tAUI =

ΔΔ

=Δ

Δ=

)(tAtA ΔΔ

Substitute (25 6) we have

Figure 25-921

21 22

022

0 ==+= BcEccBEcIμ

εμ

ε

Substitute (25-6), we have

The Energy in a Beam of Light)1025(

22 00

−μμ

To calculate the average intensity, we only need to replace the E and B with their rms value in the above formula.

Example 25-4

A garage is illuminated by a single light bulb. If the bulb radiate light uniformly in all directions, and consumes an average electrical power of 50.0 W. Assume that 5.00% of the electric power consumed by the bulb is converted to light. What are

(a)The average intensity of the bulb at a distance of 1.00 m from the bulb, andbulb, and

(b) the rms values of E and B at a distance of 1.00 m from the bulb?

Active Example 25-1

A ll l it li d i l b f li ht 1 00 i di tA small laser emits a cylindrical beam of light 1.00 mm in diameter with average power of 5.00 mW. Find the maximum values of E and B in the beam.

SSummary

1) Electromagnetic waves are transverse waves and it consists of two vertical components: electric filed and magnetic fieldof two vertical components: electric filed and magnetic field.

2) Energy in Electromagnetic Waves) gy g